Abaixo está uma instrução que descreve uma tarefa de geração de SQL, combinada com uma entrada que fornece mais contexto sobre os esquemas de tabela disponíveis. Escreva o código SQL que responda adequadamente à solicitação
 ###Instrução: Nomeie os pontos para quando os pontos contra forem de 450.
 ###Tabelas: CREATE TABLE 2009_2010_table (2009_2010_table_id AUTO_INCREMENT PRIMARY KEY, Club VARCHAR, Played VARCHAR, Drawn VARCHAR, Lost VARCHAR, Points for VARCHAR, Points against VARCHAR, Tries for VARCHAR, Tries against VARCHAR, Try bonus VARCHAR, Losing bonus VARCHAR, Points VARCHAR) CREATE TABLE feral_animals_introduced_species (feral_animals_introduced_species_id AUTO_INCREMENT PRIMARY KEY, Name VARCHAR, Species VARCHAR, Overview VARCHAR, Introduced VARCHAR, Feral VARCHAR, Threat level VARCHAR) CREATE TABLE regular_season_record (regular_season_record_id AUTO_INCREMENT PRIMARY KEY, Season VARCHAR, GP INT, W (OT/SO) VARCHAR, L (OT/SO) VARCHAR, Pts INT, Pts/GP VARCHAR, GF – GA VARCHAR, Rank (league/conference) VARCHAR, Top Scorer VARCHAR) CREATE TABLE current (current_id AUTO_INCREMENT PRIMARY KEY, Club VARCHAR, Nickname VARCHAR, Location VARCHAR, Home Ground VARCHAR, GFL Premierships VARCHAR, Years in GFL VARCHAR) CREATE TABLE night_series_winners_1977–1987 (night_series_winners_1977–1987_id AUTO_INCREMENT PRIMARY KEY, Year INT, Winners VARCHAR, Grand Finalist VARCHAR, Scores VARCHAR, Venue VARCHAR, Crowd INT, Margin INT, Season Result VARCHAR)  
SELECT Points for FROM 2009_2010_table WHERE Points against = 450